Calculus of parametric equations (finding surface area)

[jrg_pz]

I was wondering what the surface area would be when the curve:

x=e^tsint,
and y=e^tcost where (t) is greater than or equal to (0) and (t) is less
or equal to pi divided by (2).
when it is revolved about
a) the x-axis
b) the y-axis (approximation with calc. (how?))

 

[amcavoy]

I was wondering what the surface area would be when the curve:

x=e^tsint,
and y=e^tcost where (t) is greater than or equal to (0) and (t) is less
or equal to pi divided by (2).
when it is revolved about
a) the x-axis
b) the y-axis (approximation with calc. (how?))

Around the x-axis you have:

[tex]\text{SA}_x=2\pi\int_{a}^{b}f(x)\left(\sqrt{1+f'(x)^2}\right)dx[/tex]

...the y-axis you have:

[tex]\text{SA}_y=2\pi\int_{a}^{b}x\left(\sqrt{1+f'(x)^2}\right)dx[/tex]

And I assume you can figure out what f(x) and dx are in terms of your parametric equations...

 

[quicknote]

I am happy to provide a response to your inquiry about the surface area of a curve defined by parametric equations. In order to find the surface area when the curve is revolved about the x-axis or y-axis, we can use the formula for surface area of revolution, which is given by:

S = 2π∫a^b y√(1 + (dy/dx)^2) dx

Where a and b are the limits of t, and dy/dx is the derivative of y with respect to x. Let's break down the steps for finding the surface area in each case:

a) When the curve is revolved about the x-axis, we can use the formula above with a=0 and b=π/2, since these are the limits given in the problem. We can also rewrite the parametric equations in terms of x and y as:

x = e^tsin(t) → y = e^tcos(t)

Taking the derivative of y with respect to x, we get:

dy/dx = (dy/dt)/(dx/dt) = cos(t)/sin(t)

Substituting this into the formula for surface area, we get:

S = 2π∫0^(π/2) e^tcos(t)√(1 + (cos(t)/sin(t))^2) dt

Using a calculator or integration techniques, we can evaluate this integral to find the surface area.

b) When the curve is revolved about the y-axis, we can use the same formula as above, but with a=0 and b=e^π/2. This is because when we revolve the curve around the y-axis, the limits of integration will be from y=0 to the maximum value of y, which is given by e^π/2. We can also rewrite the parametric equations in terms of x and y as:

x = e^tsin(t) → y = e^tcos(t)

Taking the derivative of y with respect to x, we get:

dy/dx = (dy/dt)/(dx/dt) = cos(t)/sin(t)

Substituting this into the formula for surface area, we get:

S = 2π∫0^(e^(π/2)) e^tcos(t)√(1 + (cos(t)/sin(t))^2) dt

Again, using a calculator